您能在纯序言中写在/3之间吗? [英] Can you write between/3 in pure prolog?

问题描述

我一直在尝试了解如何从Prolog谓词中产生一系列回溯值.内置谓词between/3将在回溯时一次生成一个范围内的所有整数，因此，如何编写该整数的示例可能会帮助我完成任务.

I've been trying to understand how to produce a series of values from a Prolog predicate on backtracking. The builtin predicate between/3 will generate all the integers in a range one at a time on backtracking, so an example of how that is written may help me with my task.

我在现有的Prolog系统中寻找实现，但是GNU Prolog的between/3实现是一个C函数，诀窍在于它调用了另一个C函数"Pl_Create_Choice_Point"，从而允许它产生附加值回溯中.

I looked for an implementation in an existing Prolog system, but the implementation of between/3 for GNU Prolog is a C function, and the trick there is that it calls another C function "Pl_Create_Choice_Point" which allows it to produce additional values on backtracking.

推荐答案

bet(N, M, K) :- N =< M, K = N.
bet(N, M, K) :- N < M, N1 is N+1, bet(N1, M, K).

实际情况:

$ swipl
?- [bet].
% bet compiled 0.00 sec, 1,064 bytes
true.

?- bet(1,5, K).
K = 1 n
K = 2 n
K = 3 n
K = 4 n
K = 5 n
false.

如果使用剪切，则可以防止最终搜索失败，并恢复/3行为之间的精确内建:

If you use a cut, you can prevent the final search failure, and recover the exact builtin between/3 behavior:

bet(N, M, K) :- N < M, K = N.
bet(N, M, K) :- N == M, !, K = N.
bet(N, M, K) :- N < M, N1 is N+1, bet(N1, M, K).

实际情况:

?- [bet].
% bet compiled 0.00 sec, 416 bytes
true.

?- between(1,5,K).
K = 1 n
K = 2 n
K = 3 n
K = 4 n
K = 5.

?- [bet].
% bet compiled 0.00 sec, 240 bytes
true.

?- bet(1,5,K).
K = 1 n
K = 2 n
K = 3 n
K = 4 n
K = 5.

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